Question

Find the coordinates of the points of trisection of the line segment joining $(-4,1)$ and $(-2,-3)$.

Answer 1

R.E.F image

$A(x,y)$ $B\left(\alpha \beta \right)$

A is mid point of $P\beta$

$\Rightarrow (x,y)=(\frac{\alpha -4}{2},\frac{\beta +1}{2})$____(1)

also B is mid of AQ

$\Rightarrow (\alpha ,\beta )=(\frac{x-2}{2},\frac{y-3}{2})$____(2)

$\beta =\frac{y-3}{2}=\frac{\frac{\beta +1}{2}-3}{2}$

$\beta =\frac{\beta +1-6}{4}$

$\alpha =p-5$

$\beta =\frac{-5}{3}$

$\alpha =\frac{x-2}{2}=\frac{\left(\frac{\alpha -4}{2}\right)-2}{2}$

From (1) and (2)

$\alpha =\frac{\alpha -4-4}{4}$

$4\alpha =\alpha -8$

$3\alpha =-8\Rightarrow \alpha =\frac{-8}{3}$

$\therefore \alpha ,\beta =(-8/3,-5/3)$

$\therefore (x,y)=(\frac{-8/3-4}{2},\frac{\frac{-5}{3}+1}{2})$

$(x,y)=(\frac{-8-12}{6},\frac{-5+3}{6})$

$(x,y)=(\frac{-20}{6},\frac{-2}{6})=(\frac{-10}{3},\frac{-1}{3})$

$\therefore A\to (\frac{-10}{3},\frac{-1}{3})$ and $B\to (\frac{-8}{3},\frac{-5}{3})$